E Board: Suggested Books     ICSE Board:  Foundation Mathematics 
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Q.1 Two angles of a triangle measure 58^{\circ} \ and\  87^{\circ} . Find the measure of the third angle.

Answer:

Sum of all the angles of a triangle = 180^{\circ} 

Let the third angle = x 

Therefore 58 + 87 + x = 180 

Or x= 35^{\circ} 

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Q.2. Three angles of a triangle are in the ratio of 2 : 5 : 8.  Find the angles of the triangle.

Answer:

Let the angles be 2x, 5x, \ and\  8x 

Therefore 2x+5x+8x = 180 

Or x= 12 

Therefore the angles are 24^{\circ}, 60^{\circ}, \ and\  96^{\circ}. 

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Q.3. One angle of the triangle is 76  and the other two angles are equal. Find the measure of each angle of the triangle.

Answer

Let the two equal angles be x^{\circ} 

Therefore 76 + 2x = 180 

Or x = 52 

Therefore the angles are 76^{\circ}, 52^{\circ} \ and\  52^{\circ}. 

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Q.4. One angle of the triangle is 85^{\circ}   and the other two angles are in the ratio 8 :11 Find the measures of each of the unknown angles.

Answer:

Let the two unknown angles be 8x \ and\  11 x 

Therefore 85+8x+11x = 180 

Or 19x = 95 

Or x = 5 

Therefore the two angles are 40^{\circ} \ and\  55^{\circ}. 

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Q.5. Three angles of a triangle measures (2x)^{\circ}, (4x-7)^{\circ} \ and\  (5x-11)^{\circ} .  Find the value of x   and the measures of all the angles.

Answer

2x + (4x-7) + (5x-11) = 180 

Or x = 18 

Hence the angles are 36^{\circ}, 65^{\circ}, \ and\  79^{\circ} 

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Q.6. In  \Delta ABC, \angle A=x^{\circ}, \angle B=(2x-15)^{\circ}, \ and\  \angle C=(3x+21)^{\circ} . Find the value of x and the measure of all the angles of the triangle.

Answer:

x+(2x-15)+(3x+21)=180 

or x=29 

Hence \angle A=29^{\circ}, \angle B=43^{\circ}, \ and\  \angle C=108^{\circ} 

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Q.7. In \Delta ABC, \angle A=(2x+15)^{\circ}, \angle B=x^{\circ}, \ and\  \angle C=(3x-15)^{\circ} . Find the value of x and the measure of all the angles of the triangle and show that the triangle is an isosceles triangle.m55x

Answer:

(2x+15)+x+(3x-15)=180 \ or\  x=30 

Hence \angle A=75^{\circ}, \angle B=30^{\circ}, \ and\  \angle C=75^{\circ}.  

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Q.9. In \Delta ABC, \angle A=3\ \angle B=6 \ \angle C  . Find the angles of the triangle.

Answer:

Let \angle A=3\angle B=6 \angle C=6x 

\Rightarrow  \angle A=6x,  \angle B=2x, and \angle C=x 

\Rightarrow  6x+2x+x=180 \ or\  x=20 

\Rightarrow   \angle A=120^{\circ}, \angle B=40^{\circ} \ and\  \angle C=20^{\circ}   

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Q.10. In the adjoining figure, it is given that \angle ABC=40^{\circ}, \angle ACD=70^{\circ}, \angle ACB=x^{\circ} \ and\  \angle EAC=y . Find the value of $latex x and y .

Answer:p10

x+70=180 \ hence\  x=110^{\circ} 

Note: Sum of angles of adjacent angles on a straight line =180 

\Rightarrow 40+110+(180-y)=180 

Note:Sum of all angles of a triangle =180 

\Rightarrow y=150^{\circ} 

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Q.11. In the adjoining figure, it is given that \angle A=50^{\circ} \ and\  \angle B=70^{\circ} and BO \ and\  CO  are bisectors of \angle B \ and\  \angle C respectively. Find the measure of \angle BOC.

Answer:P11

Let \angle OCA= \angle OCB=x^{\circ}

Since BO \ bisects \angle A=ABC, \angle ABO= \angle OBC=35^{\circ}.

\Rightarrow 50+70+2x=180

\Rightarrow  x=30^{\circ}=\angle OCB

\Rightarrow 35+30+ \angle BOC=180

\Rightarrow  \angle BOC=115^{\circ}

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Q.12. Using the information given in the adjoining figure, find the value of x \ and\  y. P12

Answer:

36+y+x=180

(24+36)+2x+64=180

\Rightarrow  x=28^{\circ}

\Rightarrow 36+y+28=180 \ or\  y=116^{\circ}

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Q.13. In the adjoining figure, DCBE is a straight line, \angle ABD=135^{\circ} \ and\  \angle ACE=125^{\circ} . Find \angle BAC.

Answer:P13

 \angle ABC=180-135=45^{\circ}

\angle ACB=180-125=55^{\circ}

Therefore \angle BAC=180- \angle ABC- \angle ACB

=180-45-55=80

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Q.14. Find, giving reasons, the unknown angles marked by x, \ y in the following figures.

Answer:P14

i) 58+x+(180-123)=180

\Rightarrow x=65^{\circ}

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ii) 42+y=180 m32x

\Rightarrow y=138^{\circ}

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iii) x+72+42=180

\Rightarrow x=68^{\circ}

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iv) 50+x+x=180

\Rightarrow x=65^{\circ}

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v) 65+y=180

\Rightarrow y=115^{\circ}

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vi) 30+y+110=180

\Rightarrow y=40^{\circ}

Given AC=AD which means \angle ACD= \angle ADC=70^{\circ}

x+70+70=180

\Rightarrow x=40^{\circ}

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Q.15. In the adjoining figure, if \angle A=(x+5)^{\circ}, \angle B=(2x+3)^{\circ} \ and\  \angle BCD=(5x-14)^{\circ} , find the value of x and hence, find the measure of   i)\  \angle BCD \ ii)\  \angle ACB

Answer:P15

\angle A+ \angle B+ \angle C=180^{\circ}

\Rightarrow (x+5)+(2x+3)+{180-(5x-14)}=180

\Rightarrow x=11 \ Therefore\  \angle A=16^{\circ}, \angle B=25^{\circ}

\Rightarrow  i) \angle BCD=41^{\circ} and ii) \angle ACB=180-41=139^{\circ}

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Q.16. In \Delta ABC , side AC has been produced to D. If \angle BCD=125^{\circ} \ and\  \angle A : \angle B=2 :3 , find the measure of \angle A \ and\  \angle B.

Answer:P16

Let \angle A=2x \ and\  \angle B=3x.

\Rightarrow 2x+3x+(180-125)=180

\Rightarrow x=25

Therefore \angle A=50^{\circ} \ and\  \angle B=75^{\circ}

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Q.17. In the adjoining figure, it is given that: BC \parallel DE,\ \angle BAC=35 \ and\  \angle BCE=102 . Find the measure of i) \angle ABC ii)\  \angle ADE \ and\  iii) \angle CED

Answer:P17

i) \angle ABC+35+(180-102)=180

\Rightarrow  \angle ABC=67^{\circ}

\angle DEA+102=180   Note:Co-interior Angles

\Rightarrow  \angle DEA=78^{\circ}

ii) \angle ADE+78+35=180

\Rightarrow  \angle ADE=67^{\circ}

iii) \angle CED+67+35=180 m58x

\Rightarrow  \angle CED=78^{\circ}

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Q.18. In the adjoining figure, it is given that:AB=AC, \angle BAC=36^{\circ}, \angle ADB=45^{\circ} and \angle AEC=40^{\circ} . Find i) \angle ABC, ii) \angle ACB iii) \angle DAB and iv) \angle EAC

Answer:P18

AB=AC \Rightarrow \angle ABC= \angle ACB=x

Therefore 36+2x=180\Rightarrow x=72

i) & ii) Hence \angle ABC= \angle ACB=72^{\circ}

iv) \angle CAE+(180-72)+40=180

\Rightarrow \angle CAE=32^{\circ}

iii) \angle DAB+(180-72)+45=180

\Rightarrow \angle DAB=27^{\circ}

\angle DAE=27+36+32=95^{\circ}

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Q.19. Using the information given in the adjoining figure, calculate the values of x \ and\  y. P19

Answer:

In \Delta ABC

34+(180-x)+62=180

\Rightarrow x=96^{\circ}

In \Delta DCE 24+96+(180-y)=180

\Rightarrow y=120^{\circ}

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Q.20. Sides AB \ and\  AC of \Delta ABC  have been produced to D and E respectively. BI and CI are bisectors of \angle CBD and \angle BCE respectively. If  \angle BAC=70^{\circ} and \angle ACB 50^{\circ} , find the measure of \angle BIC P20

Answer:

Given \angle DBI= \angle IBC=x and \angle BCI= \angle ICE=y

\therefore 50+2y=180\Rightarrow y=65

\angle ABC+50+70=180

\Rightarrow \angle ABC=60^{\circ}

60+2x=180

\Rightarrow x=60

\therefore 60+ \angle BIC+65=180

\Rightarrow \angle BIC=55^{\circ}     

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Q.21. In the adjoining figure, it is being given that BC \parallel DE,  \angle CED=70^{\circ}, \angle CBA=84^{\circ} \ and\  \angle BAC=x^{\circ} .  Find the value of x .

Answer:P21

\angle DEC= \angle BCA=70^{\circ} Note:Alternate angles

x+84+70=180 \Rightarrow x=26^{\circ}

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Q.22. In the adjoining figure, it is being given that BC \parallel DE, \angle EDC=25^{\circ}, \angle ECD=20^{\circ}, \angle ABC=70^{\circ}  \angle BAC=x^{\circ} and \angle DEA=y^{\circ} . Find the value of x \ and\  y .

Answer:P22

In \Delta DEC,

25+20+ \angle DEC=180

Rightarrow  \angle DEC=135^{\circ}

\therefore y+135=180

\Rightarrow y=45^{\circ}

 BC \parallel DE \Rightarrow   \angle CBD= \angle EDA=70^{\circ}

In \Delta ADE,

70+45+x=180

\Rightarrow x=65^{\circ}

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Q.23. In the adjoining figure, it is given that \angle ACB=42^{\circ}, \angle DEB= \angle CAB=90^{\circ}, \angle ABC=x^{\circ}, \angle BDE=y^{\circ} \ and\  \angle AFE=z^{\circ}. Find the value of x, \ y \ and\  z.

Answer:

In $latex \DeP23lta FEC, 90+42+(180-z)=180 &s=1$

\Rightarrow z=132^{\circ}

In \Delta ADF, (180-132)+90+y=180

\Rightarrow y=42^{\circ}

In \Delta DBE, 42+90+x=180

\Rightarrow x=48^{\circ} 

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Q.24. In the adjoining figure, AE \parallel BC, \angle DAE=x^{\circ}, \angle ACB=(x-15)^{\circ}, \angle BAC=(\frac{x}{2}+y)^{\circ} \ and\  \angle ABC=(y+15)^{\circ}

Answer:P24

In \Delta ABC, (y+15)+(x-15)+(\frac{x}{2}+y)=180

\Rightarrow 2y+\frac{3x}{2}=180 ... ... ... ...(i)

 Since AE \parallel BC,

x=(y+15) ... ... ... ...ii) Note:Alternate Angles

Substituting ii) in i) we get

2y+\frac{3}{2} (y+15)=180

\Rightarrow 4y+3y+45=360\Rightarrow y=45^{\circ}

Substituting the value of y \ in \ i) x=45+15=60^{\circ} 

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Q.25. In the adjoining figure, BA \parallel CD . Find the values of x \ and\  y. P25

Answer:

In \Delta ABC

(y+7)+(y-10)+(4x-5)=180

\Rightarrow 2y+4x=188 ... ... ... ... (i)

Since BA \parallel CD,

(y-10)=5x ... ... ... ...(ii)

Substituting ii) in i) we get

2(5x+10)+4x=188

\Rightarrow 14x=168

\Rightarrow x=   12^{\circ} 

Substituting in ii) we get     y=70^{\circ}

E Board: Suggested Books     ICSE Board:  Foundation Mathematics 
Class 8: Reference Books     Class 8: NTSE Preparation 
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